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Momentum Conservation

  1. Jan 4, 2006 #1
    Hello, I do not know how to do this momentum conservation problem in the vertical direction, when the object is to find how far the second object went.
    A gun is fired vertically into a 1.40-kg block of wood at rest directly above it. If the bullet has a mass of 21.0g (.021 kg) and a speed of 210 m/s, how high will the block rise into the air after the bullet becomes embedded in it?
    I have found momentum which was 44.1 kgm/s and the velocity of the block which was 3.15 m/s but I just do not know how I would find the distance the block traveled. The answer is .491 m. Quick help is appreciated, thank you.
     
  2. jcsd
  3. Jan 4, 2006 #2

    Tide

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    HINT: After the collision, mechanical energy is conserved.
     
  4. Jan 4, 2006 #3
    I'm still a little confused, I've tried doing mgh = 1/2mv^2 + mgh but I realized I do not have the starting height...I am lost to what else I am supposed to do regarding this problem.
     
  5. Jan 4, 2006 #4
    Just define your coordinate system so that the starting height of the block (i.e. the height of the table, or whatever it's on) is zero. Then when you solve for h that will be the height it reached off the table.
     
  6. Jan 4, 2006 #5
    Thank you, I have solved this problem, it was really just a simple interpretation problem on my part, thanks again, I have one more question I do not quite understand that I expect to see on my exam tomorrow, I will post it in a new thread called "Momentum Particles" if anyone is kind enough to help me with it before the end of tonight.
     
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